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Let $t_{1}, t_{2},\dots$ be a real numbers such that $t_{1}+t_{2}+\dots+t_{n}=2n^{2}+9n+13$, for every positive integers $n\geq2$.If $t_{k}=103$ , then $k$ equals

Given that,
Let, $t_{1}, t_{2}, \dots$ be a real numbers.
And,

• $t_{1} + t_{2}+ \dots + t_{n} = 2n^{2} + 9n + 13 \quad \longrightarrow (1)$
• $t_{1} + t_{2} + \dots + t_{n-1} = 2(n-1)^{2} + 9(n-1) + 13 \quad \longrightarrow (2)$

From the equation $(1),$ subtract the equation $(2),$ we get

$(t_{1} + t_{2} + \dots + t_{n-1} + t_{n}) – ( t_{1} + t_{2} + \dots + t_{n-1}) = 2n^{2} + 9n +13 – [ 2(n-1)^{2} + 9(n-1) +13]$

$\Rightarrow t_{n} = 2n^{2} + 9n +13 – [ 2(n^{2}+ 1 -2n) +9n – 9 + 13]$

$\Rightarrow t_{n} = \require{cancel} {\color{Red} {\cancel{2n^{2}}}} + {\color{Teal} {\cancel{9n}}} + {\color{Blue} {\cancel{13}}} – {\color{Red} {\cancel{2n^{2}}}} – 2 + 4n – {\color{Teal} {\cancel{9n}}} + 9 – {\color{Blue} {\cancel{13}}}$

$\Rightarrow t_{n} = 4n + 7 \quad \longrightarrow (3)$

We have, $t_{k} = 103$

From the equation $(3),$ we get

$t_{k} = 4k + 7$

$\Rightarrow 103 = 4k + 7$

$\Rightarrow 4k = 96$

$\Rightarrow \boxed {k = 24}$

Correct Answer $: 24$

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